MHT CET · Maths · Continuity and Differentiability
The function defined by
\(
f(x)=\left\{\begin{array}{cc}
\frac{x-4}{|x-4|}+a \text { if } x<4 \
a+b \text { if } x=4 \
\frac{x-4}{|x-4|}+b
\end{array}\right.
\) \(\text { if } x>4\)
is continuous at \(x=4\), are
- A \(a=0, b=1\)
- B \(\mathrm{a}=1, \mathrm{~b}=0\)
- C \(\mathrm{a}=1, \mathrm{~b}=-1\)
- D \(\mathrm{a}=-1, \mathrm{~b}=0\)
Answer & Solution
Correct Answer
(C) \(\mathrm{a}=1, \mathrm{~b}=-1\)
Step-by-step Solution
Detailed explanation
L.H.L at \(x=4\)
\(\lim _{h \rightarrow 0} \frac{4-h-4}{|4-h-4|}+a=\lim _{h \rightarrow 0} \frac{-h}{|-h|}+a=\lim _{h \rightarrow 0}\) \(\frac{-h}{h}+a=-1+a\)
R.H.L at \(x=4\)
\(\lim _{h \rightarrow 0} \frac{4+h-4}{|4+h-4|}+b=\lim _{h \rightarrow 0} \frac{h}{|h|}+b=\lim _{h \rightarrow 0} \frac{h}{h}\) \(+~b=1+b\)
For continuity at \(\mathrm{x}=4\)
\(-1+\mathrm{a}=\mathrm{a}+\mathrm{b}=1+\mathrm{b} \Rightarrow \mathrm{a}=1\) and \(\mathrm{b}=-1\)
\(\lim _{h \rightarrow 0} \frac{4-h-4}{|4-h-4|}+a=\lim _{h \rightarrow 0} \frac{-h}{|-h|}+a=\lim _{h \rightarrow 0}\) \(\frac{-h}{h}+a=-1+a\)
R.H.L at \(x=4\)
\(\lim _{h \rightarrow 0} \frac{4+h-4}{|4+h-4|}+b=\lim _{h \rightarrow 0} \frac{h}{|h|}+b=\lim _{h \rightarrow 0} \frac{h}{h}\) \(+~b=1+b\)
For continuity at \(\mathrm{x}=4\)
\(-1+\mathrm{a}=\mathrm{a}+\mathrm{b}=1+\mathrm{b} \Rightarrow \mathrm{a}=1\) and \(\mathrm{b}=-1\)
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